## Particality

### Discover Fundamental Physics

##### SHARE ON

$$E=M C^2$$ .

Probably the most famous – and widely known – scientific equation of all. It implies “the equivalence of energy and mass,” is often said. But what does that mean? To cut through to the substance, we need to examine both sides, and the context in which the equation stands.

The seemingly easiest bit is $$C$$, the speed of light, which is very precisely known. In fact, it can be measured so precisely that the international system of units defines the speed of light to be precisely 299,792,458 metres per second, and uses this to define the metre in terms of the second. This is because methods for directly measuring length are less accurate.

Actually the fact that the speed of light is a constant  is not obvious. For example, if a moving object emits a light ray parellel to its motion – think of the headlight of a moving car – wouldn’t the light get a little extra speed from this? In fact the constancy of the speed of light is so totally counter-intuitive that it was, originally, a big obstacle both for the discovery and the acceptance of relativity theory.

We can read the equation as a conversion formula, somewhat similarly to what we would do to convert pounds to dollars. For example, we can calculate that one kilogram of mass gives about 90,000,000,000,000,000 joules ($$9 \times 10^{16}$$J) of energy, equivalent to 21.5 megatons of TNT equivalent. Or we can read the formula from right to left and calculate that a 15 megaton bomb uses up a mere 698g of mass to unleash all its destructive power – out of its several hundred kilograms of nuclear fuel.  Note that when a nuclear bomb is actually detonated, some very violent things are going on: nuclei are split, radiation is produced, which then heats air, etc. Similarly, but less violently, in a nuclear reactor. Mass and energy are emphatically not the same, in any generic sense. But the mass of something (here, the fissile material) can be converted into the energy of something else (here, radiation). In the present example, mass  behaves like as a particular form of stored energy, like the energy stored in a battery, which by means of a light bulb or LED can also be converted into radiation.  The convertibility also works the other way around.

In fact, in particle physics we convert energy into mass all the time, and vice versa. In some cases, the conversion of mass to energy is complete. The most powerful particle collider in the world is the LHC at CERN.  It collides pairs of protons with a collision energy of 13  teraelectronvolts or 13 TeV ($$13 \times 10^{12}$$ eV),  or 2.1 microjoules ($$2.1 \times 10^{-6}$$ J). The proton has a mass of $$1.67 \times 10^{-27}$$ kg. We can try to use $$E = M C^2$$ to convert the mass of two protons to energy:

$$2 \times 1.67 \times 10^{-27} {\rm kg} \times C^2 = 3.0 \times 10^{-10} {\rm J}$$

This is a far cry from 13 TeV. The reason is that the most famous version of Einstein’s formula fails to count the kinetic energy. When something is moving, the correct formula is

$$E=\sqrt{M^2 C^4+P^2 C^2}$$.     (1)

$$P$$ is  the momentum of the proton or, for that matter, any system, whether particle or plane or planet. The momentum is zero if something is not moving, in which case $$E=\sqrt{M^2C^4 + 0^2 C^2} = M C^2,$$  our famous starting point. The protons in the LHC ring move with a momentum of $$P = 3.47 \times 10^{-15}$$ kg m/s. This may seem like a small number, but  is a lot for an elementary particle; the protons in fact move very close to the speed of light. You can check for yourself that plugging this in the formula (1) gives 6.5 TeV (equal to 1.05 microjoules). The second proton moves equally fast in the opposite direction, giving the same magnitude of momentum and contributing another 6.5 TeV.

Every once in a while a collision will produce a heavier particle, such as a Higgs boson.  Because energy is conserved, the LHC can produce particles of a mass of at most $$13 \,{\rm TeV}/C^2$$. For this limiting case to happen, all the energy would have to go into the mass of the new particle, with the new particle being produced at rest. Most of the time, multiple particles are produced, and a large part of the collision energy becomes kinetic energy of the produced particles. Generally, the lighter a particle that might be produced, the more likely this is to happen; working out the probabilities requires substantial expertise and effort. For example, the Standard Model of particle physics predicts that very roughly one million Higgs particles have been produced in 2016 so far.

Once the particle is produced it typically decays straight away into lighter particles, which may hit and cause a signal in a detector. For example, the particle might decay into two photons, as shown in the animation below, and the energies of the two photons are what is measured.

Equation (1) is all that is needed to work out the mass of the heavy particle. This works as follows. Because energy is conserved, the two photons’ energies add up to the energy of the heavy particle (which is not directly observed). Similarly the momenta of the photons sum to the momentum of the particle. Because photons have zero mass, equation (1) applied to each photon implies $$E_1 = P_1 C$$ and $$E_2 = P_2 C$$. Finally, substituting all this into equation (1) as applied to the heavy particle, one obtains

$$E_1 + E_2 = \sqrt{M^2 C^4 + (E_1 – E_2)^2} . (2)$$

The minus sign  accounts for the photons moving in opposite directions. (Note:  I’m  assuming that when the Higgs decays, the two photons are respectively emitted parallel and anti-parallel to the direction in which the Higgs was moving, for simplicity. The general case involves the three components of the momentum vectors, requiring a bit more algebra.) Equation (2) can be solved for the heavy particle’s mass $$M$$. Voila.

Whenever two photons emerge from  an LHC collision (a “diphoton”, in the parlance of our times), $$M$$ can be calculated from (2). Tabulating all the diphoton events as a histogram might look like in the following one, obtained by the ATLAS experiment at the LHC:

The bump in the middle of the figure is the Higgs particle, it sits on top of a huge “background” of two-photon observations which are not due to production and decay of a Higgs. Such non-Higgs diphotons  can have any mass and are not expected to show a peak at any particular mass value. The bump provided an important piece of the evidence for the Higgs particle discovery in 2012, together with evidence for Higgs particles decaying in different ways, and evidence taken by another LHC experiment, CMS.

The bump in the figure is at 126.5 GeV, whereas the current best determination, combining more data, is 125 GeV. The latter value is used in the animation: You can see this by moving the velocity slider all the way to the left, such that the decaying Higgs particle is not moving.

The error bars are due to a statistical uncertainty, essentially due to the fact that the quantum mechanics underlying all particle physics predicts only probabilities. Every once in a while, statistical fluctuations can give the fake impression of a “particle bump.” This appears to have happened in the measurements shown in the following figure, taken in 2015.

This plot is suggestive of a new particle with mass of about 750 GeV and caused much excitement and activity in the particle theory community. Unlike the Higgs boson, such a particle had not been expected to exist, and its discovery would have required us to change the laws of physics at the most fundamental level. Alas, adding further events observed in 2016 to the histograms, the bump fades away.

To return to our starting point, the deeper meaning of Einstein’s famous equation is really that it gives us a definition of mass: An object’s mass is its total energy when at rest, up to a constant factor of $$C^2$$. This holds for any object or system, whether small or large, elementary or composite. In the simple example above, this energy (of the decaying particle) is measurable by converting it (letting it decay) into radiation (photons) first.

There are other ways of measuring energy and/or momentum which involve the general case (1), which  tells us in addition how the energy increases if the object is moving.  Notice that there is a fundamental difference between an object’s mass and its energy: The energy of an object depends on the object’s motion (or more precisely, the relative velocity between the object and the observer). The full content of $$E = M C^2$$ and its general case (1) is to say that every object has a mass, which, together with the speed of light, is relativistically or “Lorentz-invariant.” (This means the particular relation between mass, energy, and momentum is as in (2).)

If $$P$$ is much smaller than $$M C$$, one can derive the approximation $$E \approx M C^2 + \frac{1}{2} \frac{P^2}{M},$$ to equation (1), although this requires a little bit of calculus (or ingenuity). The first term is a constant and the second term is equal to the nonrelativistic kinetic energy. (Alternatively, one can use $$P = M V$$ to rewrite the kinetic energy as  $$\frac{1}{2} M V^2$$, which may be more familiar to the reader.)

In other words, the mass in (1) and (2) really is identical to the familiar concept of (inertial) mass of (nonrelativistic) mechanics. In the words of the master, “the inertia of a body depends on its energy content” (originally phrased as a question, in the title of his 1905 paper).

-->